## Wednesday, July 2, 2014

### GTO Brainteaser #6 Solution: Barreling and Bluffing

Today I'm going to go through the solution to the multistreet barreling / bluffing game that I introduced in the 6th brainteaser.  Learning how to solve this game illustrates most of the key concepts of multistreet GTO theory (these concepts also apply to single streets with multiple rounds of betting/raising), but there are a bunch of technical details around the concept of subgame perfection and backwards induction that I am going to gloss over here as they aren't super relevant.  I go through this game in more depth in one of my CardRunners videos "The Theory of Winning Part 3" which should be released sometime in the next few weeks.

The game was structured as follows:

There are 2 players on the turn, and the pot has 100 chips, effective stacks are 150 chips.  The Hero has a range that contains 50% nuts and 50% air and is out of position.  The Villain has a range that contains 100% medium strength hands that beat the Hero's air hands and lose to his nut hands.

For simplicity, assume that the river card will never improve either players hand.  You can also assume that the Hero is first to act (it turns out this doesn't actually matter).

The key question was, what is the EV of betting 50 chips on the turn and then having to option to bet 100 chips on the river with optimal ranges if our opponent calls optimally.  How does that compare to the EV of shoving the turn for 150 chips if the hero shoves an optimal range and the villain calls optimally.

In this post I am going to go through the math a bit quickly because I am mostly focused on demonstrating the technique of using backwards induction to solve these types of games.  For those interested, Keepitsimple had some great questions around clarifying some of the math in the comments which I answered in detail.

### Backwards Induction

The basic way to solve these types of games is to work your way backwards.  You first want to solve the river with an arbitrary hand range and then that will tell you the EV for actions on the turn that put you into any possible river state.  I explained the concept of backwards induction here.  The basic idea is that you solve the later stages of the game (in this case the river) and determine the EV of playing that game when both players play optimally.  You then consider reaching that part of the game tree  as just immediately giving you the fixed EV payoff of optimal play.

What makes applying backwards induction directly here a bit tricky is that the distribution of nuts to air that the betting player will end up with on the river depends on the turn strategy he employs and his EV of the river game depends on his hand distribution at the start of the river.  This is similar to how the distribution of hands that you raise determines the starting distribution of hands that you have when considering a 4-bet.

What this means is that we have to solve a parameterized version of the river game that will tell us the EV of reaching the river with an arbitrary hand distribution before we can backwards induct to the turn.

### Solving the River

Solving this river component of this game is quite simple.  Lets start by writing out the EV equations for the betting player when he holds air after bluffing the turn.  Call c, the frequency with which his opponent is calling a bet.

EV[bluff] = 200 * (1-c) - 100c
EV[give up] = 0

Indifference conditions require that any mixed strategy equilibrium has the same EV for bluffing and giving up.  Which implies:

200 * (1-c) - 100c = 0
c = 2/3

Similarly lets write the EV for the calling player.  Call a the probability that the betting player is bluffing when he bets.

EV[call] = 300 a + -100(1-a)
EV[fold] = 0

Again, a mixed strategy equilibrium requires these are equal which means:

a = 1/4

Clearly the betting player should always bet the nuts, and then he should bet enough air such that 1/4th of his betting range is air.  The calling player should call 2/3rds of the time.  There are two pure strategy equilibrium cases.  One occurs when the bettor has no nuts in his range, in which place he should never bluff and the calling player should always call.  The other occurs when more than 3/4ths of the bettors range is the nuts, in which case he always bets and the calling player should always fold.

Given these strategies, what is the EV of reaching the river for each player?

For the calling player he is indifferent between calling and folding when facing a bet so his EV is 0 when he is bet to as that is the EV of folding.  When his opponent checks and gives up with his air the calling player always wins the pot, so his overall EV for the river game is just the pot times how often he is checked to.  Call x the portion of the betting players range that is the nuts at the start of the river.  The actual value of x will be driven by the optimal turn strategies.  As we saw above, the betting player will always bet the nuts and will bet enough air that 1/4th of his betting range is air.  This means he will bet x + x/3 = 4x/3 of the time for x <= 3/4 and always for x > 3/4.

Thus, the calling players EV is then 200 * (1 - 4x/3) when x < 3/4 and 0 otherwise.

The betting players EV for reaching the river with a range that contains x nuts is then 200 * 4x/3 when x < 3/4, and 200 otherwise since the EVs must sum to the pot.

### Back to the Turn

Using the results above we can now write out the EV equations for various actions on the turn.  Lets start with the calling player.  The EV of calling a 50 chip turn bet when our opponent is betting a range that is z% nuts is our expectation on the river minus the cost of a call:

EV[call] = 200(1 - 4z/3) - 50
EV[fold] = 0

These are equal when z = 9/16.  If the bettor always bets the nuts then he should bet his air 7/9ths on the turn because we start with 1/2 nuts, 1/2 air and (1/2) / (1/2 + 1/2 * 7/9) = 9/16 .  This means that on the river, the betting player has the nuts x = 9/16 which is less than 3/4 and puts us in our mixed equilibrium case.

For the betting player, his EV when betting with air is very simple because the EV of reaching the river with air is 0 when x < 3/4.  Call c the probability that the calling player calls.

EV[bluff] = 100 * (1-c)  - 50c
EV[give up] = 0

So again indifference requires that c = 2/3.

### Our Solution

To summarize our GTO strategies, the betting player always bets the nuts on both streets.  On the turn he bets his air with probability 7/9, and on the river he bets with his air with probability 3/7 (since 9/16ths of his range is the nuts which he always bets and we want him to bet a range that is 1/4th air we need 3/7*7/16 to get 3/16 air).  The calling player calls in all situations with probability 2/3.

When both players follow these strategies the EV for the betting player is simple to calculate

EV[when we have air] = 0 by indifference
EV[when we have the nuts] = 100 * 1/3 (they fold the turn) + 150 * (2/3 * 1/3) (they call the turn and fold the river + 250 * (2/3 * 2/3) (they call the turn and the river) = 1600/9

Overall the betting players EV is thus 1/2(0) + 1/2 (1600/9) = 800/9

And since the game is 0 sum the calling players EV is 100/9

Along the way we assumed that the betting player always bets the nuts.  Were he to check the nuts it would be optimal for him to shove the river with his nuts and some % of his air.  As we'll see next, his EV with the nuts is lower in the game where he shoves, so deviating to checking the nuts is not profitable and this is an equilibrium.

### What about shoving the turn

If we were to bet 150 on the turn lets write the EV equations for the betting and the calling player, again calling c the probability that the calling player calls, and a the probability that the betting player is bluffing when he bets.

EV[calling] = 250 * a - 150 * (1-a)
EV[folding] = 0

Indifference requires that a = 150/400 = 3/8

EV[betting air] = 100 (1-c) - 150c
EV[giving up] = 0

Indifference reques that c = 100/250 = 2/5

The betting players EV when both players play these strategies is again simple:

EV[air] = indifference = 0
EV[nuts] = 100 * (3/5) + 250 (2/5) = 160

Overall EV = 1/2(0) + 1/2(80) = 80

So betting half pot twice is 800/9 - 80 = 80/9 chips higher in EV than jamming.

### Where does this extra EV come from?

The extra EV comes from an effect that I call "Compounding the nuts".  The basic idea is that every time we bet an optimal polarized range our opponent has to treat the air component of that range as if it were nuts.  The air in that range is effectively converted to the nuts.  If we are on the turn and know that when we reach the river some of our range will be converted to the nuts, we can play the turn as though that portion of our range already is the nuts and thus bluff with a higher frequency.

This is a very powerful idea and its effect is that in situations like this where we have a polarized range, every additional street that we can use to bet increases our EV by letting us bluff wider and wider.

For those of you who are interested, I go over the concept of compounding the nuts in detail in my CardRunners video "The Theory of Winning Part 3" which should be released soon.